Question: What is the curve described by
$$x=t^2+t+1, \, y=t^2-t+1$$
My method: Differentiating both $x$ and $y$ wrt $t$,
$$dx/dt=2t+1.$$ $$dy/dt=2t-1.$$ $$dx/dy=(2t+1)/(2t-1).$$ $$x-y=2t.$$ $$dx/dy=(x-y+1)/(x-y-1).$$
Cross multiplying, $$xdx-ydx-dx=xdy-ydy+dy.$$ Integrating, $$x^2/2+y^2/2-xy-x-y=c.$$ I don't know how to identify this curve. It isn't a parabola, ellipse or hyperbola because the coefficients of $x^2$ and $y^2$ are the same. How do I identify the curve?
Hint. Eliminate the parameter $t=(x-y)/2$ from $x=t^2+t+1$, then you will obtain immediately the equation: $$x^2-2xy+y^2-2x-2y+4=0.$$ Let $X=\frac{1}{\sqrt{2}}(x-y)$ and $Y=\frac{1}{\sqrt{2}}(x+y)$ (this is a $45^{\circ}$-rotation) then the equation becomes $$Y=\frac{X^2+2}{\sqrt{2}}.$$ What kind of curve is this?